Invertible Integers under Multiplication/Corollary 1
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Corollary to Invertible Integers under Multiplication
Let $a, b \in \Z$ such that $a b = 1$.
Then $a = b = \pm 1$.
Proof
Let $a b = 1$.
From Integer Multiplication Identity is One, $1$ is the identity element for $\Z$.
Then, by definition, both $a$ and $b$ are invertible elements of $\Z$ for multiplication.
So by Invertible Integers under Multiplication, $a = \pm 1$ and $b = \pm 1$.
As $-1 \times 1 = -1$, it follows that either:
- $a = b = 1$
or
- $a = b = -1$.
$\blacksquare$
Sources
- 1982: P.M. Cohn: Algebra Volume 1 (2nd ed.) ... (previous) ... (next): Chapter $2$: Integers and natural numbers: $\S 2.2$: Divisibility and factorization in $\mathbf Z$